剑指offer | 二叉树中和为某一值的路径 | 30

目录

• 思路分析

思路分析

就是DFS,然后到达叶子结点的时候就判断是否满足条件,如果满足的话就是一个答案.

cpp

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:

    vector<vector<int>> ans;
    vector<int> path;

    void dfs(TreeNode* root,int sum){
        if(!root)return;
        sum -= root->val;
        path.push_back(root->val);
        if(!root->left && !root->right && !sum)ans.push_back(path);
        dfs(root->left,sum);
        dfs(root->right,sum);
        path.pop_back();


    }

    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        dfs(root,sum);
        return ans;
    }
};

python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
        ans = []
        path = []

        def dfs(root,sum):
            if not root:return
            sum -= root.val
            path.append(root.val)
            if not root.left and not root.right and not sum:
                ans.append(list(path))
            dfs(root.left,sum)
            dfs(root.right,sum)
            path.pop()
        
        dfs(root,sum)
        return ans